Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const double EPS = 1e-7, PI = acos(-1.0);
using namespace std;
template <class T>
inline T IN(T& num) {
  num = 0;
  char c = getchar(), f = 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = 1, c = getchar();
  while ('0' <= c && c <= '9') num = num * 10 + c - '0', c = getchar();
  if (f) num = -num;
  return num;
}
const int NUM = 100010;
const int MAXV = NUM, MAXE = NUM * 2;
struct edge {
  int next, to;
} e[MAXE];
int head[MAXV], tot;
void gInit() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void add_edge(int u, int v) {
  e[tot] = (edge){head[u], v};
  head[u] = tot++;
}
int n, m, q;
int c[MAXV];
int sz[MAXV], fa[MAXV], dep[MAXV], start[MAXV], belong[MAXV], id[MAXV],
    len[MAXV], total;
int dfn[MAXV];
bool vis[MAXV];
void split() {
  int top = 0, tail = 0;
  dep[dfn[top++] = 1] = 0;
  fa[1] = 0;
  while (tail < top) {
    int u = dfn[tail++];
    for (int i = head[u]; ~i; i = e[i].next) {
      if (e[i].to != fa[u]) {
        dep[dfn[top++] = e[i].to] = dep[u] + 1;
        fa[e[i].to] = u;
      }
    }
  }
  memset(vis, 0, sizeof(vis));
  total = 0;
  while (top) {
    int u = dfn[--top], v = -1;
    sz[u] = 1;
    for (int i = head[u]; ~i; i = e[i].next) {
      if (vis[e[i].to]) {
        sz[u] += sz[e[i].to];
        if (v == -1 || sz[e[i].to] > sz[v]) v = e[i].to;
      }
    }
    if (v == -1) {
      id[u] = len[++total] = 1;
      belong[start[total] = u] = total;
    } else {
      id[u] = ++len[belong[u] = belong[v]];
      start[belong[u]] = u;
    }
    vis[u] = true;
  }
}
vector<int> Merge(vector<int> a, vector<int> b) {
  vector<int> res;
  for (int i = 0, j = 0, k = 0; k < 10; ++k) {
    if (i == (int)a.size() && j == (int)b.size())
      break;
    else if (j == (int)b.size() || (i != (int)a.size() && a[i] < b[j]))
      res.push_back(a[i++]);
    else
      res.push_back(b[j++]);
  }
  return res;
}
struct SegmentTree {
  vector<vector<int> > v;
  int n;
  void init(int N) {
    n = N + 5;
    v.clear();
    v.resize(n * 4);
  }
  void insert(int p, int l, int r, int x, int val) {
    v[p].push_back(val);
    if ((int)v[p].size() > 10) {
      sort(v[p].begin(), v[p].end());
      v[p].erase(v[p].end() - 1);
    }
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (x <= mid)
      insert((p << 1), l, mid, x, val);
    else
      insert((p << 1 | 1), mid + 1, r, x, val);
  }
  vector<int> query(int p, int l, int r, int L, int R) {
    if (L <= l && r <= R) return v[p];
    int mid = (l + r) >> 1;
    if (L <= mid && R > mid) {
      vector<int> a = query((p << 1), l, mid, L, R);
      vector<int> b = query((p << 1 | 1), mid + 1, r, L, R);
      return Merge(a, b);
    } else if (L <= mid)
      return query((p << 1), l, mid, L, R);
    else
      return query((p << 1 | 1), mid + 1, r, L, R);
  }
} sg[MAXV];
vector<int> Query(int u, int v) {
  vector<int> res;
  int x = belong[u], y = belong[v];
  while (x != y) {
    int& w = dep[start[x]] > dep[start[y]] ? u : v;
    int& z = dep[start[x]] > dep[start[y]] ? x : y;
    res = Merge(res, sg[z].query(1, 1, sg[z].n, id[w], len[z]));
    w = fa[start[z]];
    z = belong[w];
  }
  u = id[u], v = id[v];
  if (u > v) swap(u, v);
  res = Merge(res, sg[x].query(1, 1, sg[x].n, u, v));
  return res;
}
int main() {
  int u, v, a;
  IN(n), IN(m), IN(q);
  gInit();
  for (int i = 1; i < n; ++i) {
    IN(u), IN(v);
    add_edge(u, v);
    add_edge(v, u);
  }
  split();
  for (int i = 1; i <= total; ++i) sg[i].init(len[i]);
  for (int i = 1; i <= m; ++i) {
    IN(a);
    u = belong[a];
    v = id[a];
    sg[u].insert(1, 1, sg[u].n, v, i);
  }
  while (q--) {
    IN(u), IN(v), IN(a);
    vector<int> ans = Query(u, v);
    a = min(a, (int)ans.size());
    printf("%d", a);
    for (int i = 0; i < a; ++i) printf(" %d", ans[i]);
    puts("");
  }
  return 0;
}